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I'm following Griffiths' intro QM text, 2nd edition. We have defined the angular momentum operator and obtained the commutation relations $[L_i, L_j] = i\hbar\epsilon_{ijk} L_k$. We notice in particular how the components are incompatible observables so there's no sense in trying to simultaneously diagonalising them. Instead, after noticing that $[L_i, L^2] = 0$, we try to simultaneously diagonalise $L^2$ and one component, say $L_z$.

We define the ladder operators $L_\pm = L_x + iL_y$, and observe that, since $[L^2, L_\pm] = 0$ and $[L_z, L_\pm] = \pm\hbar L_\pm$, if $\psi$ is a simultaneous eigenstate of $L^2, L_z$ with corresponding eigenvalues $\lambda,\mu$, then $L_\pm\psi$ is also a simultaneous eigenstate of $L^2, L_z$ with corresponding eigenvalues $\lambda, \mu\pm\hbar$.

Thus if one simultaneous eigenstate $\psi$ is known, from it we can obtain a whole sequence of simultaneous eigenstates by repeated application of the ladder operators. We observe however that for any given eigenstate $\lambda > \mu^2$, so that this method of construction of new states must fail at some point. We conclude that there must exist a "top" state $\psi_+$ (or "bottom" state $\psi_-$) such that $L_\pm\psi_\pm$, the result of raising/lowering it, cannot be normalisable.

At this point, Griffiths starts an argument to relate the eigenvalues $\lambda$ and $\mu$ (the conclusion is that $\mu_\pm = \pm\hbar l$ for some positive integer or half-integer $l$, and $\lambda=\hbar^2 l(l+1)$). The first argument presented relies on the assumption that $L_\pm \psi_\pm = 0$, which as we remarked earlier is not necessary—$L_\pm\psi_\pm$ need only be non-normalisable.

QUESTION: How to obtain the result without this unjustified assumption $L_\pm\psi_\pm = 0$? i.e. either justify it or assume only non-normalisability.

A footnote in the page mentions the fact that the assumption $L_\pm \psi_\pm = 0$ is not completely due, and refers the reader to Problem 4.18, which is said to explore this. Using

$$ L_\pm L_\mp = L^2 - L_z^2 \pm \hbar L_z \qquad\qquad L_\pm^\dagger = L_\mp $$

as suggested, I can obtain

$$ |L_\pm\psi|^2 = \<L_\pm\psi|L_\pm\psi\> = \lambda - \mu(\mu\pm\hbar) $$

but at this point Griffiths seems to assume $\lambda=\hbar^2 l(l+1)$ where $\mu_\pm=\pm\hbar l$ (which can then be used to conclude that $|L_\pm\psi_\pm|=0$ indeed), but that is cheating since these had been previously derived under the assumption we're trying to avoid.


2 Answers 2


I am not sure to understand what you mean by non-normalisable. However, the only non-normalisable vector in a vector space with scalar product is the zero vector. So what it is actually obtained is that $L_\pm \psi_\pm =0$, and this should close the apparent gap you pointed out and should conclude the proof.


It is an amazing coincidence that I was revisiting that point in Griffths book today. The reason for it was replacing $L$ by $J = L + S$. As $J$ follows the same commutting relations than $L$, I thought that the conclusions should be similar, explaining also the half integers of the solution. I am still struggling with the maths, but here is what I got to this point:

Supposing $f$ eigenfunction of $J^2$ and $J_Z$ and normalized, and let's examine the conditions for $J_+f$ also be normalized: $<fJ_+^†|J_+f> = <fJ_-|J_+f> = <f|(J_x^2 + J_y^2 + i(J_xJ_y - J_yJ_x))|f> = <f|(J_x^2 + J_y^2 - J_z)|f> = <f|(J^2 - J_z^2 – J_z)|f> = 1$

$J^2f = \lambda f$ and $J_z^2f = J_zJ_zf = \mu^2f$ so: $\lambda – \mu^2 – \mu = 1$; $\mu^2 + \mu – \lambda + 1 = 0$

$\mu = -1/2 +/-(1/4 – (1 – \lambda))^{1/2} = -1/2 +/- (\lambda - 3/4)^{1/2}$

Now, the minimum value for \lambda is $3/4$, because the eigenvalues are real. So, $\mu = -1/2$. That means: for that lambda, only a $f$ with $\mu = -1/2$ can be raised, (if the functions are normalized).

The same can be done for $J_-f$, and $f$ must have $\mu = 1/2$ to be lowered.

So, for $\lambda = 3/4$, $\mu = -1/2$ or $1/2$.

If we want to lower $\mu$ to $-3/2$, the consequence from the conditions of normalization (the equation relating $\lambda$ and $\mu$) is that $\lambda = 7/4$.

And if we use $\lambda = 7/4$ for $J_-f$, $f$ must have $\mu = 3/2$ to be lowered.

The next $\lambda = 19/4$

That expression $(\lambda - 3/4)^{1/2}$ assumes the values of $l$ $(0, 1, 2 ...)$.

$3/4$ is certainly the eigenvalue for $S^2 = S(S+1) = (1/2)(1/2+1)$

So,$(\lambda - 3/4)^{1/2}$ = $(J^2 - S^2)^{1/2} = L$

But I can not find the eigenvalue for $L^2 = l(l+1)$. Neither a general formula for the eingenvalues of $J^2 = \lambda$.

Edit from Abr, 27th:

The procedure above led me to nowhere. But after trying to understand to origin of $L^2 = l(l+1)$, only through the ladder procedure, i realized that it is impossible.

That eigenvalue results from the solution of the angular part of the Schrodinger equation for a spherical symmetric potential. That differential equation happens to be the same as the generated by:

$$L^2f = (L^+L^- + L_z^2 - L_z)f$$

The spherical harmonics that solve the resultant differential equation requires that:

$$(L^+L^- + L_z^2 - L_z)f = l(l+1)f$$

where $l$ is a non negative integer.


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